 So, now this figure is of not much significance for us, I think I am not going to write this derivation, I will just go through the steps step by step because it is every step is there here and we are not skipped even plus and minus. So, I would strongly encourage slash recommend you to sit down and do this derivation on paper and pen when you go in the evening home back. So, let us write the equation this is what is the temperature now is a function of space only x that is thickness in case of calculator let us say I said thickness that is it is only thickness, but in the other directions that is in this direction and in this direction it is infinite, it is infinite. So, now I have del square t by del x square equal to 1 upon alpha del t by del t how did I write this equation our mother that is the heat diffusion equation this has come from heat diffusion equation. So, what are the boundary conditions t at x equal to 0 of t equal to t s that means as I said I have started flowing the hot water that is how I have maintained constant wall temperature boundary condition. So, t at x equal to 0 of t equal to t s that is if I have to tell what is x equal to 0 in this problem this is x equal to 0 and this is x equal to m. So, t at 0 comma t that is at any point of time right from beginning to the last my surface temperature that is at x equal to 0 is going to be t s at, but as x tends to infinity this infinity need not be infinity that is all that I am saying is for l even we can take l, but I am just putting it as infinity even if x tending to infinity at any point of time this temperature is going to be t i that is it is going to be maintained at initial temperature that is this is at equal to l I said all that I am saying is for soils this x is tending to infinity x is in this direction tending to infinity that is t i the initial temperature and at any x at time equal to 0 that is initially it is t i. So, we have it is a second order different second order in x and single order in time. So, we need two boundary conditions and one initial condition. So, that is what we have specified here we have two boundary conditions and one initial condition. Now, if I have to get this is a partial differential equation getting a solution of partial differential equation is quite difficult. So, we need to convert this partial differential equation into ordinary differential equation how do we do that I am going to do that with a small circus that is I will take eta equal to x upon square root of 4 alpha t how on earth one knows that I have to define eta equal to x upon square root of 4 alpha t and suspending that question for two minutes I will answer that through scale analysis little later, but for now you take my words and take eta as x upon square root of 4 alpha t. Let us take x upon square root of 4 alpha t alpha is my thermal diffusivity t is my time x is the location of my temperature where I am interested in. So, del square t by del x square equal to 1 upon alpha del t by del t eta equal to x upon square root of 4 alpha t. So, for del t by del t I can write this as del t by d t by d eta into del eta by del t that is I have just multiplied you can imagine that d eta d eta I have multiplied. So, minus d t by d eta into if I differentiate eta with respect to t that is x will remain as it is x upon 2 t square root of 4 alpha t. This is simple differentiation I think you can all catch along with me. So, similarly I need del square t by del x square. So, that is del t by del x I can write it as d t by d eta equal into del eta by del x. You please note the careful subtle difference whenever I am writing differentiation of t with respect to eta I am using ordinary derivative whenever I am using differentiation of eta with respect to either t or x I am using partial derivative. Why because we will realize later that when I use eta my temperature is going to become independent of x and t it is going to be a only a function of eta. So, that is how I can transform p d e into o d e that is partial differential equation into ordinary differential equation. So, del t by del x equal to d t by d eta into del eta by del x equal to d t by d eta into 1 upon square root of 4 alpha t this I got by differentiating eta with respect to x. Again if I differentiate this d t by del t by del x with respect to x I get del square t by del x square that is d by d eta into del t by del x into del eta by del x that is 1 upon square root of 4 alpha t I have to multiply with this square root of 4 alpha t I end up getting 1 by 4 alpha t into d square t by d eta square. So, now I am successful in transforming this equation which is in terms of x and t in two in terms of eta. So, if I substitute this in this equation so I get I substitute whatever I have found here in del square t by del x square and 1 by alpha del t by del t if I substitute and simplify I am going to get d square t by d eta square equal to minus 2 eta d t by d eta. So, now you see now the partial differential equation has got itself converted into ordinary differential equation earlier my temperature was function of x and t now it is only a function of eta this solving this equation becomes very easy. Before I solve that I let us take care of boundary conditions and initial condition how do they get transformed that is now I need two boundary conditions that is as x equal to 0 my t I equal to 0 that is at x equal to 0 that means eta equal to 0 at eta equal to 0 my temperature is going to become t s at eta equal to 0 temperature is t s at x t x tending to infinity that is eta tending to infinity t is t i these two boundary conditions are sufficient for me to solve this equation. So, now coming back coming back I had left one question earlier that is how on earth one knows that eta equal to x upon square root of 4 alpha t it has just not just like that it has not come it has come with a logic or a reasoning that reasoning is what I am going to tell you. So, now let us just take simple egg which is what has been drawn here if this egg is hot enough it is taken directly sorry this is let us take a copper ball sorry it is not egg because egg I have to cook it and take it out and it starts getting cooled but here I have shown the temperature where in which the temperature is increasing it is egg only initial solid temperature is decreasing that is an egg which is cooked in a cooker has taken out has been taken out what will happen only the outer layer will start to decrease in its temperature inside very much that is what is called yellow yellow thing what is that called yolk. Yoke will perhaps not feel only the all women which has got boiled will start feeling that the temperature dip. So, that is what is being shown here. So, the core region that is the yolk and the all women region this is the outer layer the outer layer only is starting feeling the dip in the temperature but the core region that is the yolk region does not know that it has been thrown out from cooker to the atmosphere. So, now having understood this this layer where in which it has been where there is a temperature gradient let me call that as a skin layer and delta. So, this delta let us let us keep this in our mind initial temperature is T i and fluid temperature that is the air temperature into which it has been brought is T infinity and surface temperature be T naught let us keep this in mind that is again there here for the sake of if you forget. So, T i is the initial solid temperature T naught is the surface temperature T infinity is the fluid temperature. Now, let us put get back to this equation del square T by del x square equal to 1 upon alpha del T by del T let me take del square T by del x square. So, first before I do that del square T by del x square let me take del T by del x at x tending to 0 or let me go little slow del square T by del x square can be written as del T by del x at x equal x of the order of delta. Please note I am writing notation tilde where in which we see in our keyboards tilde is there first one first in the top we see tilde just below escape button tilde is there that tilde that means it is of the order of when I say order of see this example I usually take in the class also when we keep a purse and if we have in our purse if we have kept money of the order of thousands if we lose our purse we are worried because we have lost thousands of rupees. Let us say I have just put 1 rupee 2 rupees 3 rupees total amounting to 10 that is tens of rupees if I lose perhaps I do not mind so much because it is only of the order of tens. So, but if I lose thousands of rupees then I will be worried because I have why do I get worried because my order of money which I have lost is too much. So, that is what I mean when I say here order of magnitude. So, so x is of tilde delta means it is of the order of delta that is until it is not equal to delta it is not equal to 10 rupees or 20 rupees it is of the order of tens of rupees that is all I mean. So, that is why here I am saying del square t by del x square is of the order of del t by del x at x of the order of delta minus del t by del x at x of the order of 0 that is very much nearer that is very much on the outer surface upon delta minus 0. This let us find now del t by del x at x tending to delta there is no temperature gradient here there is no temperature gradient. So, I can put that as 0, but what about del t by del x at x tending to 0 that is very much on the outer surface there is a slope you can see that slope what is that slope t i minus t naught upon delta minus 0. Now, I do not think any confusion after yesterday's tutorial problem why how did I write this. So, this is t i minus t 0 upon delta is my del t by del x at x tending to 0 or x of the order of 0. If I substitute this here in del square t by del x square I will get minus of t i minus t naught upon delta square. So, now let us keep this you know in the back of our mind. Now, let us come to del t by del t del t by del t is t naught minus t i upon t minus 0 initially it was t i that was when time is 0, but after passage of time t it has become t naught. So, that is t naught minus t i upon t. So, what do I get now if I substitute these orders of magnitudes here I get t i minus t naught upon delta square is of the order of I am not using the equal to sign please note this tilde I am using 1 upon alpha t naught minus t i upon t. So, if I absorb this minus here I get t naught minus t i both the numerators vanish. So, then I get delta square is of the order of alpha t. So, that is delta is of the order of alpha t. So, that is what is my eta, eta equal to delta upon square root of alpha t that is x upon square root of alpha t. This four is just a mathematical convenience because if I take four most of my equations in this derivation become easier even I can derive the solution without taking the four there is no problem, but now I have answered the question why we have defined eta as x upon square root of alpha t. So, that is what comes from order of magnitude analysis delta is of the order of alpha t. So, only within this medium I am taking only within the layer delta we are taking it as semi infinite medium very much away it does not know that I have thrown the egg does not know that or the yolk does not know that I have thrown out from cooker to atmosphere. So, with this now it is mathematics I tend to go fast when it is mathematics whenever there is physics only I tend to go slow. So, I am now going to go fast and d square t by d eta square equal to minus 2 eta d t by d eta if I integrate this I am going to end up with this. So, again if I integrate this I get log of d eta d t by d eta equal to minus eta square plus c prime and again if I integrate this I am going to get this equation if I substitute these boundary conditions I am going to get t minus t s upon t i minus t s equal to 2 upon square root of pi into all of this function. I think most of you are familiar with this function that is 0 to eta e to the power of minus eta square d eta this is nothing, but error function what we have studied in mathematics. So, this is whole of this 2 by square root of pi into all of this is called error function that means at every eta this has been tabulated if you see this this has been tabulated error function of w e 2 by square root of pi 0 to w e to the power of minus w square d v this is this has been tabulated and this is what is error function w. So, this is what is called as Gaussian error function it has been for every w there is a typo here this is w square and d w there is a typo this is d w w square and d w. So, for every w I can get the error function. So, I do not have to bother about integrating all the time it is a standard integral and it has been integrated and kept. So, you can see that for 0.1 I get 0.1, 1, 2, 4, 6. So, directly I can plug in and get my temperature distribution. So, that is T s is the surface temperature I know, T i is the initial temperature I know and T at any x I can compute what is that x how do I get at any x and at any time T I can compute. How do I get that because eta is known eta is equal to x upon square root of 4 alpha t. So, now taking this temperature distribution again our motherly equation that is Fourier's law of conduction if I put heat flux equal to minus k d t by d x at x equal to 0 equal to minus k into T i minus T s. If I differentiate this all of this mathematics if you do I am not going to do that for you if you do that I am going to get k into T s minus T i upon square root of pi alpha t. I know consciously I have skipped the steps here in this transparency which is integration and here which is differentiation I would strongly argue to do this mathematics yourself because it becomes too lengthy for me to go through we will be unnecessarily losing time in algebra rather than heat transfer. So, I would not like to lose time in algebra. So, we have got the temperature distribution we have got the heat flux. So, with this we can work out like this here for constant heat flux for constant for convective boundary condition things have been worked out, but remember one thing I want to emphasize this solution is valid for semi infinite medium that is that medium for which for the complete span of time the bottom surface does not know whatever boundary condition what we have applied on the top. What we have solved is for constant wall temperature it can be constant heat flux it can be convective boundary condition no matter what is the boundary condition applied on the top surface the bottom surface does not know what is happening on the top surface for that duration of time I can apply this solution what we have derived so far. Why is it so, so much sought after because we can get an analytical solution we can get a closed form solution why we could get a closed form solution because of this simplification I could convert the ODE into PDE that sorry PDE into ODE that is the beauty of semi infinite medium because I can get a closed form solution I go through this assumption and apply it and there is a problem on soil itself Professor Arun will take this problem see if why it is so nice this problem because I need to know there are two situations I put underground pipes I put underground pipes below the soil. Now if someone asks Bombay municipal corporation asks the question how below I need to dig and put my pipes why because of course in Bombay there is no problem in Bombay the only problem I can visualize is what happens if on the outer surface a fire is caught a lorry two lorries collide and fire has occurred and that fire should not go the heat of the fire should not go and rupture the underground pipes in a hot country in a cold country snow is being formed on the top surface and that snow or the coldness should not seep into the pipe and the water which is flowing in the pipe and supplying to various household areas it should not convert itself into ice then I will not get water supply. So to answer these questions we can take the recourse of semi infinite medium and solve and come up with this length below which I need to lay the pipe that problem will be solved by Professor Arun after the lunch break I think I will just quickly take one question or two questions perhaps. So I am going to Mufakum Jha college Hyderabad if you have any questions please ask them. Yeah in slide number 18 you have referred like transient heat conduction in large large plane walls you have referred you have not referred both stands for lambda n like x is x is a dimensionless quantity and alpha k is thermal conductivity like for lambda n you have not referred what did it refers for. Yeah see if you remind remember the question asked is x is non dimensionalized as x by l and time is non dimensionalized as alpha t by l squared but what is this lambda n what is this non dimensional number it is not a non dimensional number lambda n tan lambda n equal to biot number if you remember for plane wall those are the roots of the equation lambda n tan lambda n equal to biot number if I have to get the first root lambda 1 tan lambda 1 equal to biot number this is a transcendental equation where in which the roots of this transcendental equation we are going to obtain because we will have multiple roots the first root would be lambda 1 so it is not a non dimensional number see let us go back see there is a question what is this lambda n so lambda n is it a non dimensional number. So lambda n is not a non dimensional number let us take recap you see here lambda n tan lambda n is equal to biot number so that is these lambda are the roots positive roots of this transcendental equation. So why the transcendent what is transcendental equation transcendental equation is that equation which has multiple roots that is all so that is first root whatever I get that is lambda 1 second root I get lambda 2 third root is lambda 3 so it is not a non dimensional number it does not have units but it has no physical. Yeah if there is no physical significance as such but the lambda 1 lambda 2 lambda 3 are dependent on biot number that is what has been tabulated here. So for each biot number the lambdas are tabulated where are they tabulated yeah so for each biot number the first root has been tabulated lambda 1 yeah yeah so these are these are the four roots yeah these zetas please read them as lambda 1 lambda 2 lambda 3 lambda 4 I hope we have answered the question. So we will now go to government college Salem. Good afternoon professor can we use transient conduction analysis to calculate the time required for solidification as it involves higher conduction convection heat transfer rate is also high. Okay so one of the participants is asking the question that can we use this transient conduction solution for solidification yesterday also we have dealt to this in length transient conduction solution is indeed used in Stefan problem but only in the solid region and only in the liquid region separately but not solid and liquid together now that this question is being asked again and again I will post this Stefan problem solution in the moodle. Okay so how we can ask I think most of you are interested in either solidification or melting there is a Stefan problem to what extent these solutions can be applied is demonstrated very nicely by professor Stefan okay Stefan does not own is not only having contribution in radiation is having contribution in melting and solidification also that is Stefan problem I will post this in the moodle professor okay over to MA NIT Bhopal I want to go to slide number 37 sir slide 37 yeah in which the problem is started the partial differential equation was there yeah yeah I want to show yeah this one so in this problem the second boundary condition is when T when extending to infinity at all the time T the value of the temperature is T i that is the initial temperature so please clarify this what boundary condition yeah see what do I mean I will come back to my whiteboard I took the solid the soil so where extending to infinity means very much at the bottom of the soil it is I said that the bottom the top boundary condition whatever I put the bottom of the soil is not feeling of course extending to infinity may not be infinity all the while I that will be demonstrated when we solve the problem for now let us say extending to infinity is 10 meter so at the 10th meter below the soil surface it does not feel the hot water flowing on the top that is what I mean that is why at extending to infinity it is continuing to have the temperature T i that is the answer ma'am for your question okay thank you sir over to Amal Jyothi sir what is the significance or difference between this Fourier number and the alpha your number is a non-dimension form then both are having the same meaning no both are see the one of the participants question is alpha and Fourier number have the same meaning no I disagree with you professor the Fourier number is non-dimensional number one alpha T by L square alpha is meter squared per second time is second so numerator I end up getting meter squared and denominator I get meter squared so Fourier number is non-dimensional number one alpha the unit of alpha in the very first class we derived and the unit of alpha is meter squared per second it is thermal diffusivity the first difference between Fourier number and alpha is Fourier number is non-dimensional alpha is dimensional so but alpha is a part of Fourier number all that we are saying is for Fourier numbers very less the q dot conducted is lower than the q dot store when can this happen either when alpha is less or time is less or L is large so alpha does not mean same as Fourier number another thing is alpha does not involve any time aspect it is a material property Fourier number takes care of the time aspect so even if you have a piece of wood and you want it has a very poor alpha value if you expose it to a particular fire or whatever for days together then definitely it is not going to behave as it would behave if it was put in the fire for 10 minutes so the time aspect is what is coming in the Fourier number alpha is a material property so I hope so because actually most of the textbooks tend to write even when we are teaching alpha itself when we define itself we say when alpha is large response is large we tend to write but we should not be looking at we should not look at alpha as a single entity we have to look alpha as embedded in Fourier number. Yeah I think in the introduction to heat transfer we say k by rho c p n we will directly put that as k is large so alpha is large it will respond fast it is not that alone and rightly pointed out actually it is a very good question very good question alpha t by l squared has to be looked Fourier number is far more significant than alpha alone a material with high alpha but with a very very large l will take more time time to respond just because it has a high thermal conductivity does not mean that it is going to respond very quickly if you are having if you have a wall made of a particular material several meters in length it is going to be different. One more question sir in a lemmed heat analysis suppose we take the example of a thermometer so when do we actually measure the temperature? So one of the participant has asked me a question how much long we have to wait if I use a thermometer ok all of us have felt this when we go to doctor he if we or we have used the thermometer ourselves at home when we are trying to measure our body temperature when we put the thermometer under our armpit typically we try to see the watch and wait for a minute or two why because it takes more time it because you see again I have to take who is responding for temperature here the bulb the mercury which is there in the bulb that is rho v c p upon h a s so you take I would like you to go back and calculate whoever has asked this question take rho c p of mercury and volume by area of the bulb portion and h you take it as 15 because it is exposed to atmosphere if you calculate it will so turn out that it will come around 60 to 65 seconds so that is why doctor or ourselves we will wait for 1 minute to 2 minutes so that we get the steady state temperature of our body that is the answer ok so PSG college we will take a question from you yeah there is somebody coming hurriedly hello sir yeah sir can you teach us the shortcuts for Cartesian coordinate to a cylindrical and spherical coordinate conversion shortcuts I think students will be able to do that for all the governing equations I really do not remember when you cannot just do it like that because r sin theta r cos theta those kind of things have to be done right coordinate transforms have to be done Cartesian to cylindrical right so I do not want to answer that because I do not think I know it correctly to tell you but why why are we interested as teachers in the shortcuts students are the ones who will be interested right maybe students will be able to answer that I have 2 students here from a neighboring college let us see do you know the shortcuts no they also do not know the shortcuts so I really cannot answer that right away but coordinate transformation is what you mean I think right so just looking at the equation you cannot how maybe I can tell you how to remember spherical I do not remember I cannot remember cylindrical I can tell you how to remember that is all that d by dz term will remain as it is and your radius term would be 1 by r d by dr of k r d t by dr because the r has to be inside because it is function of radius and the other one because it is the d square t by d theta square so theta is dimensionless so you will have 1 by r square sitting so 1 by r square d by d theta and the bracketed term that is what I remember for cylinder spherical coordinates I actually do not remember it so I do not think I would be able to answer the question the way you have asked ask any of the students they will be able to tell you and if you get back if you come across the shortcut let us know okay let us get back to this semi infinite body semi infinite solid and let us look at the example problem this was very elegantly done the delta is of the order of under root alpha times t was very elegantly done and the solution was presented in terms of an error function so of course students do not have to remember any of this thing as long as they understand the concept of a semi infinite solid and if you are given sufficient time and the resources I think all of them can even derive this it is not a big problem okay so solution is of the nature of error function and the heat flux or the heat transfer rate is just as we did in the film once I get the temperature distribution so let us appreciate this aspect our aim always has been to get the temperature distribution first because a simple differentiation of the temperature distribution is going to give me the heat flux okay so minus k dt by dx at x is equal to 0 is going to give me the heat flux at that location and that is constant because we are talking of heat flux so that is just a differentiation of the error function and those are also tabulated you get q double prime is k ps minus pi divided by under root pi alpha t so that in fact professor Prabhu had mentioned this factor of 4 was introduced here you know in the similarity variable of eta under root 4 alpha t eta is equal to x over under root 4 alpha t if this 4 was not there you would you would keep having some square root terms in the argument of the error function so here right now the error function it is a nice function here you know this 2 by root pi comes out as it is that is why this factor of 4 was introduced otherwise you would have to multiply and divide by some new some number later on to get it into this particular form because the definition of error function is of this kind or the factor of 4 was introduced so historically probably that is what is the reason anyway so there are such equations derived for constant wall temperature which is what we studied mostly we have similar thing for constant wall heat flux and constant surface convection boundary condition and there you see there is a complementary error function of argument eta which is nothing but 1 minus the error function of eta so you go to the error function table so you can get the complementary error function of course it is a little bit messy in terms of the arguments etcetera but nevertheless it is a doable exercise so we will go to this numerical problem in areas where air temperature remains below 0 degree centigrade for a sufficiently long periods of time freezing of water in underground pipes is a major concern and of course this happens very much in cold countries and in cold places even here in India so that is a definite no no because first of all water when it freezes it is going to expand because of anomalous behavior beyond below 4 degrees that is one thing then you will not have water for consumption so both these reasons and in cold countries especially Scandinavian countries it is so cold for about 4 to 6 months of the year so you cannot afford to have pipes freezing so this is a problem with very relevant practical application soil remains relatively warm during these periods and it takes weeks for the sub freezing temperatures to reach the water mains in the ground thus soil effectively serves as an insulation to protect the water from the freezing temperature ok so now that is the history let us look at the problem the ground at a particular location is covered with snow back at minus 10 degree centigrade for a continuous period of 3 months so surface temperature boundary condition problem you have T s is minus 10 and the time period is given 3 months means 90 days and you can obviously convert into minutes and seconds so on and so forth the average soil properties are k given alpha is given to us assuming initial uniform surface temperature of 15 degree centigrade for the ground determine the minimum burial length to prevent water pipes from freezing so 15 degree centigrade is the ground temperature is given and there was a heavy snowfall and it is covered the ground completely you are asked to find how long meaning how much deep the pipe has to be so that for 3 months definitely it is not going to be affected roughly from December to February let us say so this is a schematic known T surface minus 10 T i is also 15 degree centigrade k is 0.4 alpha is 0.15 times 10 to the minus 6 meter square per second so with this we can it is a very straight forward problem nothing to worry temperature in the soil is affected by thermal conditions at one surface only therefore the soil can be considered as a semi infinite medium the earth is one of the best possible examples of a sink of semi infinite medium etc so whatever you do you abuse the earth here on the surface nothing is felt in the core it takes years after years to for the earth to respond in the form of a volcano or earthquake but what you do is going to be very very slow to reach the center thermal properties are given and we have the only relation that we have is this one so we have a temperature we need to calculate the distance x at which the pipe has to be buried so T is 0 degree how did we get T is 0 degree water this is at 0 degree so T is 0 so at after 3 months at that location temperature should be 0 degrees so anything beyond this x location anything further in depth if we bury we are going to be safe even after 3 months that is what it means so 0 minus of minus 10 divided by 15 is the initial temperature minus of minus 10 how did I get that all this is given quantity surface is minus 10 initial temperature is 15 degree centigrade that is equal to error function of x over under root 4 alpha T x is what is to be determined so this quantity is known so just substituting the numbers I get x is equal to 0.8 meters so we need to have the pipes buried at least 80 centimeters 80 centimeters is hardly 2 and a half feet so 2 and a half feet if you buried for 3 months even if you have complete snow at minus 10 degree centigrade water is not going to freeze that is actually a very it is not very deep actually 2 and a half feet is hardly half the normal if you take 5 feet or 160 centimeters as the normal human being side half the depth that is it so 3 or 4 steps is what we are talking about so 80 centimeters in depth if you bury it the surface temperature is minus 10 it is going to take more than 3 months for the water to reach 0 degree centigrade so that is a simple example to illustrate the use of this formula this concept of semi infinite solids so what do you take home from this we take home 2 very important things that any object even if it is a very good conductor can be treated as a semi infinite solid for very small instance of time so if you just expose a surface even though it is a metallic surface to a change in the boundary condition and just after that immediately after that for a very small amount of time that effect is not going to propagate very much in depth you can treat that surface that object as a semi infinite solid for the analysis second thing is this concept of really large surfaces etcetera where conductivity is very very low in that case definitely the change in the boundary condition is not going to affect the other end that much so both these situations can be tackled by this concept of semi infinite solids so you can reserve some questions later on the last and I think this hardly anybody covers but since we have included this will spend about 10 minutes on this concept and then you can take a look at the problem associated with it all this while we looked at transient conduction in single dimension so x y or z whatever it was a radial direction we looked at conduction we did not we did not bother about conduction in the other dimension that is why remember in one of the earlier slides here see we looked at the concept of plane wall in case of a plane wall we said it is large plane wall when we said it is a long cylinder and we called this as a sphere of course we did not have to worry about it so large plane wall long cylinder specifically we emphasize on these two because we wanted conduction to be predominantly in one particular direction that is here the x direction in the place of plane wall we did not want any conduction aspects related to the y direction or the z direction we did not we did not want conduction in the z direction in case of a cylinder now many realized situations you do not have actual single dimensional conduction all these approximations that we did d t by d r or d t by d x is much greater than d t by d z let us just look at this thing if this temperature distribution is a function of time but this can also be treated like a function of space where y and x direction temperature is changing so definitely I will have second dimensional effect if the cylinder or if the plane wall is not very large in the other direction that way the gradients in the other direction will also become significant therefore what has to be done so best thing is you do not solve this but that is not an option so we say how do we solve this kind of problems these kind of problems are taken care by superposition of solutions of two specific geometry we call this as a product solution and what we did for whatever we have studied one actually things follow logic very beautifully actually you know we looked at plane wall if I have a plane wall or a very very long cylinder and the long cylinder is intersected by plane wall in the middle then the solid that is formed after that intersection let me take a look here the solid that is formed after the intersection is nothing but a finite short cylinder basically so if you want to we can draw this and show but it is little difficult to imagine if I have a very long infinitely long surface with a finite thickness and if I am if I am cutting it by another surface the solid which is obtained by the intersection of these two is going to be a solid of finite dimension so let me try to draw this let us see how successful I am this is what I am trying to say so this is a plane wall for example and what we are saying is this is my relevant direction of heat transfer this is y direction we have said it is a very large plane wall we do not care but if I say in three dimensions if it is like this if this is going to be cut by another plane wall of a corresponding center line here and this is my let us say this is my y direction so this intersection is going to give you a rectangular box of some finite dimensions is that correct so this box is what I will get so this box is not infinite in either direction and for such a box I will have to deal with multi-dimensional both x and y and for this box which is bounded by this surface here this surface here and this front face correspondingly the other surfaces I have to take into account the solution mathematically I am saying I have to take into account the solution of the problem for a plane wall which we have done like this for another plane wall which we have we can do by this method horizontal plane wall whatever be the nature of solution is the same whether it is horizontal or vertical and then we say superposition of these two is going to give you the solution of the actual problem which is multi-dimensional in nature so that is what is given in the slides here let me so transient temperature charts presented can be used to determine temperature distribution and heat transfer in 1D heat conduction problems associated with blah blah blah using a superposition principle called the product solution these charts can also be used to construct solution for two-dimensional transient heat conduction problem encountered in geometry such as a short cylinder long rectangular bar semi-infinite cylinder or a plate or even three-dimensional problems that are associated such as a rectangular prism semi-infinite rectangular bar so on and so forth provided all the surfaces of the solid are subjected to convection of the same fluid and at the same temperature with the same heat transfer coefficient and the body involves no heat generation why because we have analytical solutions graphical solutions whatever for only those conditions now if I impose a different boundary condition or suddenly have a heat generation my associated governing equation itself is going to change so the nature of the solution will be changed we are not going to deal with those aspects so the solution in such multi-dimensional geometries is expressed as a product of solutions for the one-dimensional geometry whose intersection yields the multi-dimensional geometry that is what I told by the intersection of these two plane wall problems. So what we say is a long cylinder long cylinder is this one where this dimension is not of concern to us but now when I say a short cylinder can of coke or any cool drink which comes in a can is a short cylinder because the dimension is about 15 centimeter and diameter this way is about 6 centimeters or so so the ratio is hardly 2 to 1 half it is not a long cylinder in this case I can say short cylinder solution you see here it is t of r comma x comma t that is what we are saying it is two-dimensional in r and x I said z we are using x here plane wall is the is what is intersecting the cylinder like this ok the long cylinder when it is cut or intersected by a plane wall gives you this short cylinder basically you are cutting whatever is the unwanted part you are sharing the top and the bottom part and getting a very nice geometry that is what it means so a plane wall solution which is of x comma t but mind you the x is like this so if I want to draw this here yeah see this here this problem which is there at the end has this a long cylinder if I do not worry about the length this is going to go very very long but if a plane wall cut plane wall of thickness 2 L cuts it like this ok. So you can imagine it like a like a case where tennis balls are there which is cut by a book for example I cannot give any better example no a case where tennis balls are there so a long cylinder or many students carry this circular box for the drawing sheets that when it is cut by a book that is probably what is going to give you a short cylinder and that what we are saying is the plane wall is like this in the vertical direction that is what is the plane wall and the radial conduction happens as it is shown here. So what is told is that for the solution we have temperature distribution non-dimensional temperature for the plane wall times non-dimensional temperature for the infinite cylinder the product of that gives you the temperature distribution of the short cylinder. Yeah there is a small problem I think again for the short cylinder only so that is what professor is saying is that it is a combination of if I take a short cylinder the solution is a combination of plane wall and infinite cylinder if someone ask me how does how do I know that it is always the product yeah it is a good question it cannot it cannot be proved actually it can be proved but it cannot be showed that always it is going to be product sometimes it can be quotient also so or it can be summation also most of the times it has so happened that the product works. So that is the reason at least in case of short cylinder plane wall and infinite cylinder solutions if you multiply the product is same as what is obtained for short cylinder so that is how these solutions have been generated because we need to go always from simple to complex like evolution so that is how we are evolving here. So we have a plane wall and an infinite cylinder so let me go to this problem which professors showed us so that is the problem says so a short brass cylinder of diameter 10 centimeter and height 12 centimeter is initially at a uniform temperature of 120 degree Celsius that is what is being shown here so that is we have a length of 12 centimeters but we are breaking it into two cylinders as professor told us now and that is how we will get from for this half I can get the plane wall and for radius I can get the cylinder solution. So that is how we make this short cylinder problem as a product of a plane wall and a long cylinder so now note here it is not L 12 centimeters but 2 L 12 centimeters that is the major difference or the major point what one should notice having noticed that initial temperature is given 120 degree Celsius height is given 12 which we have broken into two cylinders and the diameter is 10 centimeters and the initial and the atmospheric temperature is 25 degree Celsius and the heat transfer coefficient is 60 watts per meter square degree Celsius so that is convective boundary condition which is what for which the solution is available if you change the boundary condition yes we do not have the solution if you ask me whether we have we have the solutions for constant heat flux no we do not have the constant heat flux boundary condition okay so we have to calculate now the temperature the center of the cylinder how do I do that so first thing is what I do is I will have to as I said earlier it is function of Fourier number Bayotte number and the location which is non dimensionalized that is what it is how I will found out find out alpha t by L square mind you L is here 0.06 so this is we have to be very cautious we need to be very careful about choice of L usually if we give this problem in our tutorial or quiz our students end up taking L so that is the general mistake all of our students tend to make so that is the reason why I am emphasizing these are considered as two cylinders okay so fine so now I have tau I have Bayotte number 1 upon Bayotte number seeing that tau and Bayotte number I am going to get this ratio so once I get that ratio I will be able to calculate the theta so that is similarly at the center of the this is for the plane wall and for center of the cylinder also I can calculate tau t by L square now you see for L square I have taken 0.05 what is this 0.05 that is coming from my diameter that is radius I have taken okay so now taking the radius I get the tau and the Bayotte number from either calculations or the chart I am going to get at r equal to 0 that is at the center of the cylinder 0.5 so at the center if I want the temperature I have to multiply what I got for plane wall and the cylinder and that ratio that product is 0.4 so if I put that with the t infinity taking the theta definition I am going to get the temperature as 63 so similarly one can work for other locations as well that I am not going to do that is at the top surface similarly you will do take the appropriate x by L and having got theta not now take the x by L and compute the individual values and multiply that you are going to get the temperature at any other location that is how one will go ahead and get the temperatures so now similarly the way we did for temperatures the way we did for products as temperatures similarly it is possible for heat transfer total heat transfer the total transient heat transfer 2 are from multidimensional thing has been found by Langston not very long back in 1982 so that is it is q upon q maximum into 1 plus q upon q maximum into 1 minus q upon q maximum of 1 so again I am saying all of this without any proof so if I have 3 bodies also I can put so this is basically the intersection it is it is just saying that it is not 1 plus 1 equal to 2 it is 1 plus into 2 into something else because there is an interaction physically that is how I can explain this so that can be computed again here that q that can be computed we know this q by q maximum for plane wall and q by q maximum for infinite cylinder and taking that relation whatever I found I had stated little while ago this is how I can compute q by q maximum for two dimensions and get this so this is how we can go for various configuration I would urge all the participants to go ahead and take the printouts of this or if you have printouts closely watch for various situations which are given here for this for cylinders and for other configurations which are stated here so that is how we can conclude the transient conduction so as professor has already stated I do not want to reemphasize anything more but for the point that we have introduced two non-dimensional numbers biot number and Fourier number biot number is hl by k and Fourier number is alpha t by l square and here k is the thermal conductivity of solid so with this I think we will say bye bye to transient conduction and we will move on to convection okay so but before we move on to convection maybe I can just take one or two questions for next five minutes I will take one or two questions yeah please ask your questions only to transient conduction if you have any questions on fins please post them on model PSG Koyambattur question please sir to to an application spatial effect solution is available numerical solution also is possible can you comment on that which one is better see always in a the question asked is for all the solutions what have been solved for transient conduction whether with respect to space and time numerical solutions are also possible yes very much numerical solutions are also possible but which one we are recommending see in a classroom what are we trying to look at we are what is that we understood in transient conduction in a nutshell I said two non-dimensional numbers are important if I just go by numerical methods I would have never come across this biot number and Fourier number having understood biot number Fourier number and some special cases where in which the closed form solutions are the analytical solutions are possible then you can move on to numerical solutions it would be a good exercise solve the same problems numerically compare the results one with the other so I would not say that numerically superior to analytical I would always say analytically superior to numerical no doubt about that okay you can always contest between experimental and numerical but not analytical versus experimental or numerical analytical is always superior to any of course that is also limited to the assumptions what you have made while deriving that if we keep that in mind if I can get an analytical solution I would always prefer an analytical solution however if I have I am teaching numerical method if you are solving a numerical problem do not go directly to transient conduction and temperature is a function of x y z and t take for those cases where in which you have derived the closed form solutions that is for example t x t as a function of x comma t for plane wall derive solve that numerically get the analytical solution put the numerical results one on one with the analytical result and then save your numerical result is closely matching with the analytical result that is how I would look at the this question okay Jane to Hyderabad any question please Yes sir question regarding the problem which we have solved in the morning session that is egg problem okay the problem was solved you using one term solution in your discussion sir and I have solved the problem with charts okay and I got the answer as 910 seconds whereas answer from the analytical solution is 865 seconds okay the question asked by one of the participants is that is that actually one participant has solved the morning x problem using charts apparently she gets a different answer compared to the closed form solution or the analytical solution which has been told and 865 see the main they cannot come different number one number two when you take charts definitely there is going to be certain amount of error and the accuracy in chart is going to be always under question in the sense not the chart is wrong it is just that the way we are reading the chart has so many round off error that is all it is otherwise if I calculate with the calculator using the closed form solution and the chart it should be coming out same so they cannot be different this 865 and 910 that minor discrepancy is because of this round off error not that they are wrong this is just because of round off errors yeah there are so actually now you are taking two things and taking a product of this so if there is a mistake in if there is a round off error in one it gets culminated into product error error in the product so it is the round off error that is all it is I do not think you need to be worried about that at all actually you have a question please go ahead semi infinite solid problem yeah we substituted a t x comma t as 0 whereas how is it sir just tell me t t problem pipe under the earth the question asked is in the semi infinite medium why the temperature that is t x comma t is taken as 0 that is given in the cell that is given in the problem itself freezing of water freezing of water so freezing we are expecting that it will it is given no the air temperature remains below 0 for prolonged periods of time the freezing of water in the underground pipes is a major concern so freezing point is 0 degree Celsius so that is the temperature which we have taken as 0 degree Celsius okay I think we need to stop questions here